#include <iostream>
using namespace std;

const int mod=998244353,N=505;
int dp[N][N];
int a[N];

long long c[505][505]; // c[i][j] --> c[5][2]==10

void init()
{
	for(int i=0; i<505; i++)
		c[i][0] = 1;
    for(int i=1; i<505; i++)
        for(int j=0; j<=i; j++)
        {
            if(j==0)
                c[i][j]=1;
            else
                c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
        }
}

int C(int x, int y)
{
	return c[x][y];
}

int main()
{
	init();
	int n; scanf("%d", &n);
	for(int i=1; i<=n; i++) scanf("%d", &a[i]);
	dp[0][0] = 1;
	int sum = 0;
	for(int i=0; i<n; i++,sum+=a[i])
	{
		// 全部放前面 
		for(int j=2; j<=sum; j++)
			dp[i+1][j+a[i+1]] = dp[i][j];
		dp[i+1][0] = dp[i][0];
		//枚举前面放多少个(后面至少放一个)
		for(int x=0; x<a[i+1]; x++)
		{
			//枚举剩余部分从哪个位置后面开始放 
			for(int k=1; k<=sum; k++)
			{
				for(int j=k+1; j<=sum; j++)
					dp[i+1][k+1+x] = (1ll*dp[i+1][k+1+x] + 1ll*dp[i][j] * C(j - k - 1 + a[i+1] - x - 1, a[i+1] - x - 1)) % mod;
					
				//j == 0 的情况 
				dp[i+1][k+1+x] = (1ll*dp[i+1][k+1+x] + 1ll*dp[i][0] * C(sum - k + a[i+1] - x - 1, a[i+1] - x - 1)) % mod;
			}
		}
	}

	int ans = 0;
	for(int i=0; i<=sum; i++) ans = (1ll*ans + dp[n][i]) % mod;
	cout << ans << '\n'; 
	return 0;
}